Good morning

I would appreciate if someone could verify the following please.

The question:

A 100kg crate is pulled across a horizontal floor by a force P that makes an
angle of 30 degrees above the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.200. What is P if the net work
done is zero.

My solution:

First determine Ff = umg. This would give me 196N. My next step is to
determine P. I use 196N and put that over cos30. My answer is 226.3N. Does
this seem reasonable.

thanks for your time

Subject: Crate being pulled horizontally
From: "TGen"
Date: 20/12/04 15:20 GMT Standard Time
Message-id: O4Cxd.4191$uj2.1531@clgrps12

Good morning

I would appreciate if someone could verify the following please.

The question:

A 100kg crate is pulled across a horizontal floor by a force P that makes an
angle of 30 degrees above the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.200. What is P if the net work
done is zero.

My solution:

First determine Ff = umg. This would give me 196N. My next step is to
determine P. I use 196N and put that over cos30. My answer is 226.3N. Does
this seem reasonable.

Nope.

You are neglecting the fact that the upwards component of the pulling force is
reducing the contact force between the box and the ground. I think you'll find
this solution needs some equations to be solved first.

I don't much like the wording of the question. If there is a force acting over
a distance then work is being done by definition so quite how net work can be
zero in this situation I'm not sure.
--
Dave Baker - Puma Race Engines (www.pumaracing.co.uk)

I assume I should be using these?

Horizontal:
Vertical:


"Dave Baker" wrote in message
...
Subject: Crate being pulled horizontally
From: "TGen"
Date: 20/12/04 15:20 GMT Standard Time
Message-id: O4Cxd.4191$uj2.1531@clgrps12

Good morning

I would appreciate if someone could verify the following please.

The question:

A 100kg crate is pulled across a horizontal floor by a force P that makes
an
angle of 30 degrees above the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.200. What is P if the net work
done is zero.

My solution:

First determine Ff = umg. This would give me 196N. My next step is to
determine P. I use 196N and put that over cos30. My answer is 226.3N. Does
this seem reasonable.

Nope.

You are neglecting the fact that the upwards component of the pulling
force is
reducing the contact force between the box and the ground. I think you'll
find
this solution needs some equations to be solved first.

I don't much like the wording of the question. If there is a force acting
over
a distance then work is being done by definition so quite how net work can
be
zero in this situation I'm not sure.
--
Dave Baker - Puma Race Engines (www.pumaracing.co.uk)

TGen wrote:
I assume I should be using these?

Don't top post.

begin 666 407946-0.png

Don't post binaries to this news group, nor any group that
does not have "binary" in the name.

As to the original question: The only sensible meaning to "no work
being done" is to assume that the box is in motion at a constant
velocity. That is, the only work being done is against friction.
So you have a frictional force, gravity, and the pulling force,
and all are in equilibrium. Draw your free body diagram, put in
what the various forces are, and you should be able to work out
the pulling force.
Socks

"TGen" wrote in message
news:O4Cxd.4191$uj2.1531@clgrps12...
Good morning

I would appreciate if someone could verify the following please.

The question:

A 100kg crate is pulled across a horizontal floor by a force P that
makes an
angle of 30 degrees above the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.200. What is P if the net
work
done is zero.

Who set that question? It is nonsensical. There is no way that a
force P acting at 30 deg to the horizontal can lead to a situation in
which no work is done. For example:

Horizontal component of P = p cos 30 deg
This is positive
Work done by this force = P x d
This is also positive.

Franz

Your mistake:
Ff is not = umg.
Instead, Ff = uN, where N is the normal force upward.
The normal force is whatever force the floor has to exert (here,
upward) to keep the crate from penetrating it.
But N is not mg here, because the vertical component of P is also
pulling upward.

If the net work done is zero, and the displacement is not zero
(obviously -- it says the crate is being pulled across a floor), then
the net force in the displacement direction must be zero.

--PD

Forgot the work done by friction, which is negative.

Negative work? Should cool off by itself.
Also just set P=0 and your done.


"PD" wrote in message
ps.com...
Forgot the work done by friction, which is negative.

"Dave Baker" wrote in message
...

I don't much like the wording of the question. If there is a force acting over
a distance then work is being done by definition so quite how net work can be
zero in this situation I'm not sure.

I suppose that it could be argued that pulling something across
a surface is not the same as the something actually moving...

But then, one should then be dealing with the coefficient of static
friction (u_s) rather than the coefficient of dynamic friction (u_d).
Then again, u_d is usually less than u_s. So the information given
cannot even provide the upper bound of the applied force. Hmmm.

"PD" wrote in message
ps.com...
[some other point]

Ah, fancy seeing you here.
"OK, now I really do have to pay attention to my business on the road.
When I return, I will have to snip a lot of stuff out and focus on
what appears to be two stopping points."

I'm still wondering about them, since 27 Oct 04.
Androcles